Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 16 - Vector Calculus - 16.8 Exercises - Page 1151: 14

Answer

$8 \pi$

Work Step by Step

The parameterization for the given surface is: $r=\lt 2 \cos t, 2 \sin t, 1 \gt \implies dr = \lt -2 \sin t , 2 \cos t , 0 \gt$ and $F(r(t))=\lt -4 \sin t , 2 \sin t , 6 \cos t \gt$ Now, $\iint_{S} curl F \cdot dS=\iint_{C} F \cdot dr= \int_{0}^{2 \pi} \lt -4 \sin t , 2 \sin t j, 6 \cos t \gt \cdot \lt -2 \sin t , 2 \cos t j, 0 \gt$ or, $=\int_{0}^{2 \pi} 8 \sin^2 t+4 \cos t \sin t dt$ or, $=\int_{0}^{2 \pi} 8 (\dfrac{1-\cos 2t}{2})+(2)(2) \cos t \sin t dt$ or, $=\int_{0}^{2 \pi} 4+2 \sin 2t -4 \cos 2t dt$ or, $=\int_{0}^{2 \pi} 2 \sin t -8 \cos 2t dt+ \int_0^{2 \pi} 4 dt$ or, $=[4t-2 \sin 2t -\cos 2t]_0^{2 \pi}$ or, $=4 [(2 \pi) -0-1-(0-0-1)]$ or, $=8 \pi$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.